Show that the truncation error of quadratic interpolation in an equidistant table is bounded by $$\frac{h^3}{9\cdot3^{0.5}}\max f''' (x)$$
I have gotten to nothing because i don't seem to understand quadratic interpolation errors unlike linear interpolation errors.
How to get the interpolation error
In the form of the Newton interpolation, the quadratic interpolation polynomial centered at $x_k$ can be written as $$ p_2(x)=f(x_{k-1})+f[x_{k-1},x_k](x-x_{k-1})+f[x_{k-1},x_k,x_{k+1}](x-x_{k-1})(x-x_k). $$ Fixing some point $z$, the cubic interpolation polynomial with this extra node can be written as $$ p_3(x)=p_2(x)+f[x_{k-1},x_k,x_{k+1},z](x-x_{k-1})(x-x_k)(x-x_{k-1}). $$ As this is exact at the nodes, $p_3(z)=f(z)$ or $$ f(z)=p_2(z)+f[x_{k-1},x_k,x_{k+1},z](z-x_{k-1})(z-x_k)(z-x_{k-1}). $$ The mean value formula for divided differences then gives $$ f[x_{k-1},x_k,x_{k+1},z]=\frac16f'''(\zeta) $$
The maximum of the error in the interval
The interpolation error for the quadratic polynomial interpolating at the points $x_{k-1},x_k,x_{k+1}$ and their tabulated values is bounded by $$ |f(x)-p_2(x)|\le\frac{1}6\max_{\zeta\in[x_{k-1},x_{k+1}]}|f'''(\zeta)||\cdot(x-x_{k-1})(x-x_k)(x-x_{k+1})| $$ If $x\in8x_{k-1},x_{k+1}]$ is parametrized as $x_k+sh$ with $s\in[-1,1]$, then this bound becomes $$ \frac{h^3}6\|f'''\|_{[x_{k-1},x_{k+1}]}\cdot |s-s^3|. $$ The local maxima of the non-constant factor in $[-1,1]$ are located at $s=\pm\frac1{\sqrt3}$, which will result in the claimed bound.
Use in table look-up may give smaller errors
If implemented as a procedure, the input $x$ would be associated to the closest table node $x_k$ with $|x-x_k|\le\frac h2$ as center node in the interpolation. But the maximizing points $s$ are outside the interval $[-\frac12,\frac12]$. To get an approximate value for the point $x=x_k+\frac1{\sqrt3}h$ from the table, one would use the interpolation polynomial based on the triple centered at $x_{k+1}$. Thus the maximum error for the more realistic procedure is at $s=\pm\frac12$.